T Distribution Calculator
The t-distribution calculator makes it easy to compute the cumulative probability associated with a t score or with a sample mean. For help in using the calculator, read the Frequently-Asked Questions or review the Sample Problems.
To learn more about Student's t-distribution, go to Stat Trek's tutorial on the t-distribution.
Note: Both the t-distribution and the standard normal distribution assume that observations are normally distributed in the population. And as sample size increases, the t distribution becomes increasingly similar to the standard normal distribution. So when would a researcher choose the t-distribution over the standard normal distribution? A common rule of thumb is to choose the t-distribution when (1) the sample size is small and/or (2) the population standard deviation is unknown.
Frequently-Asked Questions
Instructions: To find the answer to a frequently-asked question, simply click on the question.
Which statistic should I use - the t-score or the mean score"?
What are degrees of freedom?
What is a standard deviation?
What is a t statistic?
What is a population mean?
What is a sample mean?
What is a probability?
What is a cumulative probability?
Sample Problems
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The Acme Chain Company claims that their chains have an average breaking
strength of 20,000 pounds, with a standard deviation of 1750 pounds. Suppose a customer
buys 14 randomly-selected chains. What is the probability that the average breaking strength
for the customer's' chains will be 19,800 pounds or less?
Solution:
One strategy would be a two-step approach:
- Compute a t statistic, assuming that the breaking strength for the customer's chains is 19,800 pounds.
- Determine the cumulative probability for that t statistic.
We will follow that strategy here. First, we compute the t statistic:
t = [ x - μ ] / [ s / sqrt( n ) ]
t = (19,800 - 20,000) / [ 1750 / sqrt(14) ]
t = ( -200 ) / [ (1750) / (3.74166) ]
t = ( -200 ) / (467.707) = -0.4276
where x is the mean breaking strength in the customer's chains, μ is the population mean, s is the standard deviation, n is the sample size, and t is the t statistic.
Now, we can determine the cumulative probability for the t statistic. We know the following:
- The t statistic is equal to -0.4276.
- The number of degrees of freedom is equal to 13. (In situations like this, the number of degrees of freedom is equal to number of observations minus 1. Hence, the number of degrees of freedom is equal to 14 - 1 or 13.)
Now, we are ready to use the T Distribution Calculator. Since we have already computed the t statistic, we select "t score" from the drop-down box. Then, we enter the t statistic (-0.4276) and the degrees of freedom (13) into the calculator, and hit the Calculate button. The calculator reports that the cumulative probability is 0.338.
Therefore, there is a 33.8% chance that the average breaking strength for the customer's chains will be no more than 19,800 pounds.
Note: The strategy that we used required us to first compute a t statistic, and then use the t-distribution Calculator to find the cumulative probability. An alternative strategy, which does not require us to compute a t statistic, would be to use the calculator in the "mean score" mode. That strategy may be a little bit easier. It is illustrated in the next example.
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Let's look one more time at the problem that we addressed above in
Example 1. This time, we will illustrate a different, easier strategy to solve
the problem.
Here, once again, is the problem: The Acme Chain Company claims that their chains have an average breaking strength of 20,000 pounds, with a standard deviation of 1750 pounds. A customer buys 14 randomly-selected chains. What is the probability that the average breaking strength for the customer's chains will be no more than 19,800 pounds?
Solution:
We know the following:
- The population mean is 20,000.
- The standard deviation is 1750.
- The sample mean, for which we want to find a cumulative probability, is 19,800.
- The number of degrees of freedom is 13. (In situations like this, the number of degrees of freedom is equal to number of observations minus 1. Hence, the number of degrees of freedom is equal to 14 - 1 or 13.)
First, we select "mean score" from the dropdown box in the t-distribution Calculator. Then, we plug our known inputs (degrees of freedom, sample mean, standard deviation, and population mean) into the t-distribution Calculator and hit the Calculate button. The calculator reports that the cumulative probability is 0.338. Thus, there is a 33.8% probability that the average breaking strength for the customer's chains will be 19,800 pounds or less.
Note: This is the same answer that we found in Example 1. However, the approach that we followed in this example may be a little bit easier than the approach that we used in the previous example, since this approach does not require manual computation of a t statistic.
- The principal at a local high school claims that the average IQ of his faculty is higher
than the actual IQ of 90% of teachers in the district.
The school board administered an IQ test to 15 randomly selected teachers at a his high school. They
found that the average IQ score was 115 with a standard deviation of 11. Assume that the principal's
claim about his faculty is correct. What would be the average IQ of a teacher in the district?
Note: In terms of cumulative probability, the notion that 90% of teachers in the district have an IQ of 115 or less implies that P(IQ≤115) is 0.90.
Solution:
We know the following:
- The cumulative probability is 0.90.
- The standard deviation is 11.
- The sample mean is 115.
- The number of degrees of freedom is 14. (In situations like this, the number of degrees of freedom is equal to number of observations minus 1. Hence, the number of degrees of freedom is equal to 15 - 1 or 14.)
First, we select "mean score" from the dropdown box in the T Distribution Calculator. Then, we plug the known inputs (cumulative probability, standard deviation, sample mean, and degrees of freedom) into the calculator and hit the Calculate button. The calculator reports that the population mean is 111.2.
Thus, if the principal's assessment of the IQ of his faculty is correct, we conclude that the average IQ of a teacher in the district is 111.2.